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Calorimetry Help

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meme:
Hi!
Wow, I love the new look of the forums - awesome! Anyway, I have some questions on calorimetry which I've tried, but I'm not sure I'm doing it right. I'd appreciate any help.

We did a lab on the heat of neutralization for an acid-base reaction. Anyway, these were my results:
[*] volume of acid (HCl) - 50mL
[*] temperature of acid - 20.9 C
[*] volume of NaOH - 50mL
[*] temperature of NaOH - 20.2 C
[*] exact molar concentration of NaOH - 1.0 mol/L
[*] maximum temperature from graph (we had to make a graph of our results) - 27.5 C[/list]

Calculations:
[*] Average initial temperature of acid + base: (20.2 + 20.9)/2 = 20.6
[*] Temperature change: 27.5 - 20.6 = 6.9
[*] Volume of final mixture: 50 + 50 = 100mL
[*]Mass of final mixture (assume density of solution is 1g/mL): (1)(100) = 100g
[*]specific heat of mixture(given): 4.18J/g C[/list]

Here's the part I'm not sure of:
[*]Heat evolved: (4.18)(100g)(6.9) = 2884.2 J
[*]Amount of OH- reacted:
  HCl + NaOH --> NaCl + H2O
  100g NaCl + H2O  x  1 mol     x  1 mol NaOH  x  1 mol OH-
                                  76.46g  x  1 mol            x  1 mol NaOH
   = 1.31 mol OH- ??
[*]Amount of H2O formed:
  1.31 mol NaOH  x  1 mol H2O
                              1 mol NaOH
      = 1.31 mol H2O ??
[*]Heat evolved per mole of H2O, (heat of neutralization)
  heat of neutralization = - Cp(H2O) x combined masses (acid + base) x change in temperature
  = - (4.18)(100g)(6.9) = -2884.2 J[/list]

I know there's something fishy going on there, please *delete me*


Other Questions:

1. A thermometer labeled as being miscalibrated at the factory – the freezing point of water reaches 1.2 ºC and the boiling point of water reads 101.2ºC at standard pressure. But, because you are in a hurry and the thermometer is the last one available, you decide to go ahead and use it in the experiment, recording the temperatures as they appear on the thermometer. How ill the use of this thermometer affect the determination of the specific heat of the metal?
So does it just affect the temperature by 1.2ºC? or does it affect it by something else? Then again, it probably wouldn't affect it at all, because change in temperature would still be the same, right?

2. The calorimeter, although a good insulator, absorbs some heat when the system is above room temperature. The consequence of this heat loss from the system is accounted for in the temperature extrapolation f the data. Describe an experiment that could be used to calibrate the calorimeter by measuring this heat loss.
I have NO idea about this one... hints, please?

Thanks again for all the help. I love you guys!

Mitch:
You did sime hocus-pocus with those calculations. Did you use solid NaOH or pure HCl for this experiment? I'm assuming you didn't count in the dilution factor. I'm still looking over your numbers to see where it might of gone wrong.



1.You got it. Good answer

2. You basically just add boiling water to the calorimeter and see how the temperature goes down with time. And just calculate it's specific heat.

Mitch:
do you know the molarity of the HCl or the NaOH?

Donaldson Tan:
Hmm..

Heat evolved is correct

Amount of OH- reacted:
I have no idea how u arrive at 100g NaCl. However, total volume of reaction mixture is 100ml and its density is assumed at 1g/ml. By Law of Mass Combination, there's no way NaCl present can be of 100g  :o

Heat of Neutralisation is expressed in J.mol-1 with reference to water. U must divide the heat evolved by the amount of water formed by neutralisation.

meme:
Wow.. you guys answered really fast! Anyway, here are the answers to your questions about my questions (hehe)

Mitch -

The HCl and the NaOH were diluted.
HCl - 1.1 M
NaOH - 1.0 M

Geodome:

Sorry about the typo, I actually checked everything a million times to take off any mistakes.. Guess I messed up. I had meant to put 100 g of NaOH + HCl (sorry again!) So are the numbers right, though? Is it 1.31 mol both H2O and OH-?

And about the heat of neutralization.. so I don't have to use the equation I wrote up there? So, basically, the answer would be heat evolved/mol of H2O:
 2884.2 J/ 1.31 = 2201.7 J/mol = 2.20 kJ/mol
Is that right?

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